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I want to show that, for odd positive integers $n_{1}$ and $n_{2}$, we have :

$\frac{n_{1}n_{2}-1}{2}$ = $\frac{n_{1}-1}{2}$ + $\frac{n_{2}-1}{2}$ mod(2)

Let $n_{1}$ = ${2m_{1} +1}$ and $n_{2}$ = ${2m_{2} +1}$

I get :

2${m_{1}m_{2}} + {m_{1}} + {m_{2}}$ $\cong$ ${m_{1} + m_{2}}$ mod(2)

My question is : how this is proving the statement ? I am not sure what to do next.

miskinalien
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1 Answers1

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You can drop the $2m_1m_2$ since working mod $2.$

coffeemath
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  • Thank you! How could I deduce from that that (-1/n) = (-1)^((n-1)/2) ? – miskinalien Dec 21 '17 at 20:57
  • @ miskinalien Am I right in interpreting $(-1/a)$ as the quadratic residue symbol? [That is, it is $+1$ if $-1$ is a residue mod $a$ and $-1$ if $-1$ is a nonresidue mod $a$] – coffeemath Dec 22 '17 at 01:15
  • If your $(-1/n)$ is meant as suggested in my last comment, note that the formula of your comment requires (in general) that $n$ be an odd prime, not just any odd positive integer. For example $(-1)^{((9-1)/2)}=+1$ would if true suggest that $-1=8$ is a residue mod $9$, but the residues mod $9$ are $0,1,4,7$ making $8$ a nonresidue in this case. But using non-primes as "denominators" in $(a/n)$ requires special care, look up Jacobi symbol. – coffeemath Dec 22 '17 at 02:10